The distances, x m, travelled to school by 140 children were recorded. The results are summarised in the table below. Distance, x m | x ≤ 200 | x ≤ 300 | x ≤ 500 | x ≤ 900 | x ≤ 1200 | x ≤ 1600 Cumulative frequency | 16 | 46 | 88 | 122 | 134 | 140
- (a)On the grid, draw a cumulative frequency graph to represent these results.[2]
- (b)Use your graph to estimate the interquartile range of the distances.[2]
- (c)Calculate estimates of the mean and standard deviation of the distances.[6]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [B1] Axes labelled 'cumulative frequency' (or cf) from 0 to at least 140 and 'distance (or d) [in] m' from 0 to at least 1600, linear scales with at least 3 values stated.
- [B1] All plotted correctly at correct upper end points (200 etc.) curve drawn accurately joined to (0, 0) (straight line segments B0) but no daylight above 140. Cf scale no less than 2 cm = 20 children.
- [M1] Accept 660 < UQ < 720 – 240 < LQ < 290. If values are outside our range, FT providing scales linear and increasing cf drawn.
- [A1] Accept correct evaluation of 660 < their UQ < 720 – 240 < their LQ < 290 with clear indication that graph has been used for at least one of 105 or 35. 440
- [B1] Frequencies 16 30 42 34 12 6. 5 or 6 correct frequency values seen. Mid-points 100 250 400 700 1050 1400
- [B1] 5 or 6 correct midpoint values seen.
- [M1] Values substituted into mean formula using their midpoints which must be in the class - condone 1 data error.
- [A1] 505 WWW
- [M1] Values substituted into variance formula using (their mean)² and their midpoints and their frequencies (including for denominator). Accept unsimplified. Condone 1 data error.
- [A1] 324 WWW
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